Laplace transform: where am I messing up?

[faust9]

Ok, using the definition of Laplace transforms to find \L\{f(t)\}

Given:
f(t)=\{^{\sin{t}, 0\le{t}<{\pi}}_{0, t\ge{\pi}}

So, this is what I did:

\L\{\sin t\}=\int^{\pi}_{0} e^{-st}\sin t dt+\int^{\infty}_{\pi} e^{-st}(0)dt

=\int^{\pi}_{0} e^{-st}\sin t dt

=\frac{-e^{-st}\sin{t}}{s}]^{\pi}_{0}+\frac{1}{s}\int^{\pi}_{0} e^{-st}\cos t dt

=\frac{-e^{-st}\sin{t}}{s}]^{\pi}_{0}+\frac{1}{s}(\frac{-e^{-st}\cos{t}}{s}]^{\pi}_{0}-\frac{1}{s}\int^{\pi}_{0} e^{-st}\sin t dt)

=\frac{-se^{-s\pi}}{s^2}+\frac{1}{s}(\frac{-1}{s}-\frac{1}{s}\L\{\sin t\})

\L\{\sin t\}(\frac {s^2+1}{s^2})=\frac{-se^{-s\pi}}{s^2}-\frac{1}{s^2}

\L\{\sin t\}=\frac {-se^{-s\pi}-1}{s^2+1}

Which I know is wrong because the Laplace for sin t should be:

\L\{\sin t\}=\frac {1}{s^2+1}

I know my limits of integration will affect the problem but the restricted limits should only add a term to the numerator e^{-s\pi} I believe.

Where did I mess up?

Thanks a lot.

[arildno]

Your mistake is in the 5.line, in the evaluation of cos(t) at the limits indicated.

[arildno]

Also, the sin(t) evaluation in the 5.line should be zero.

[faust9]

Got it. Thanks. How silly of me.